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If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\vec{r}$ and $\vec{\tau}$ be the torque of this force about the origin, then
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Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$
Key Idea Torque is an axial yector ie, its direction is always perpendicular to the plane containing vectors $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{F}}$.
$$
\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}
$$
Torque is perpendicular to both $\vec{r}$ and $\vec{F}$
$$
\therefore \quad \begin{aligned}
\vec{\tau} \cdot \overrightarrow{\mathrm{r}} & =0 \\
\overrightarrow{\mathrm{F}} \cdot \vec{\tau} & =0
\end{aligned}
$$
$$
\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}
$$
Torque is perpendicular to both $\vec{r}$ and $\vec{F}$
$$
\therefore \quad \begin{aligned}
\vec{\tau} \cdot \overrightarrow{\mathrm{r}} & =0 \\
\overrightarrow{\mathrm{F}} \cdot \vec{\tau} & =0
\end{aligned}
$$
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